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$a_n,b_n\in\mathbb{C}$. Is it enough that both $\sum_{n\in\mathbb{Z}}a_n$ and $\sum_{n\in\mathbb{Z}}b_n$ converge?

It seems to me that this should be enough, because: $$\sum_{k\in\mathbb{Z}}(a_k+b_k)=\lim_{n\rightarrow+\infty}\sum_{k=-n}^n(a_k+b_k)=\lim_{n\rightarrow+\infty}\sum_{k=-n}^n(a_k+b_k)=\lim_{n\rightarrow+\infty}\sum_{k=-n}^na_k+\sum_{k=-n}^nb_k$$ Now it's just $$\lim x_n+y_n=\lim x_n+\lim y_n$$ which is true if the right hand side makes sense. Is my assumption correct?

Arthur
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Tom83B
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    Note that $\sum_{n \in \Bbb Z}$ and $\lim_{k \to \infty}\sum_{n = -k}^k$ are not the same thing. – Arthur Jan 23 '17 at 11:04
  • I had no idea. Could you please elaborate further? – Tom83B Jan 23 '17 at 11:05
  • $\sum_{k \in \mathbb Z}$ is not necessarily the same thing as $\lim_{n \to \infty} \sum_{k=-n}^{n}$. Actually $\sum_{k \in \mathbb Z} = \lim_{m,n \to \infty} \sum_{k=-n}^{m}$, and this subtle point causes some trouble with your argument. – Antonio Vargas Jan 23 '17 at 11:06
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    Take, for instance, the sequence $a_n = 1/n$ (and $a_0 = 0$ or something). Then $\lim_{k \to \infty}\sum_{n = -k}^k a_n = 0$ while $\sum_{n \in \Bbb Z}a_n$ doesn't exist. Intuitively, $\sum_{n \in \Bbb Z}$ exists if the ratio between how the lower and upper limit go to their respective infinities doesn't affect anything.. – Arthur Jan 23 '17 at 11:09
  • Ok, but if $\sum_{n\in\mathbb{Z}}$ exists, then it's equal to $\lim_{n\rightarrow\infty}\sum_{k=-n}^n$ isn't it? The equality should still hold then, right? – Tom83B Jan 23 '17 at 11:15
  • @Tom83B Yes, and no. My point is just that it doesn't go the other way around. Your argument has to take into account that for $\sum_{n \in \Bbb Z} (a_n+b_n)$ to exist, the limit has to exist no matter how the lower and upper bound goes to infinity, and you have to make full use of the fact that it is true for $\sum_{n \in \Bbb Z}a_n$ and $\sum_{n \in \Bbb Z}b_n$ – Arthur Jan 23 '17 at 11:19

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